Surface Normals

Let us see about surface normals. A surface normal also called the simple normal, to a flat surface is a vector that is vertical to that surface.A line normal to a flat, the normal factor of the force, the normal vector, etc. The concept of routine generalizes to orthogonality.A normal is used for computer graphics.

Calculating a surface normals

For a convex polygon such as a triangle, a surface usual will be calculated as the vector cross product of two (non-parallel) edges of the polygon.
For a plane given by the equation ax + by + cz + d = 0.
For a plane is represent the following equation r = a + αb + βc.
where a is a vector to get onto the level surface and b and c are non-parallel vectors lying on the plane, the normal to the plane defined is given by b × c .
For a hyperplane in n+1 dimensions equation given from following equation r = a₀ + α₁a₁ + α₂a₂ + ... + α_na_n,
where a₀ is a vector to get onto the hyperplane and a_i for i = 1, ... , n are non-parallel vectors two-faced on the hyperplane, the normal to the hyperplane can be approximated by (AA^T + bb^T) ^{− 1}b where A = [a₁, a₂, ... , a_n] and b is a random vector in the space not in the linear span of a_i.

Cross product partial derivation


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In surface normal,a surface S is given completely   as the set of points (x,y,z) it satisfying F(x,y,z) = 0, so the normal point (x,y,z) on the plane.

For a surface S given clearly as a function f(x,y) of the independent variables x,y (e.g., f(x,y) = a₀₀ + a₀₁y + a₁₀x + a₁₁xy).

The normals can be identified in at least two equivalent ways.

The first getting its contained from F(x,y,z) = z − f(x,y) = 0, from the normal follows readily.

Angle Sum Theorem

In a triangle the sum of all the three interior angles will be equal to 180^o.  If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles. Here we are going to see about the angle sum theorem.

Understanding Angle Bisector Theorem is always challenging for me but thanks to all math help websites to help me out.

Angle sum theorem:

Angle sum theorem of a triangle is equal to two right angles, i.e., 180 degrees
Given:
ABC is a triangle
To Prove
Angle A + Angle B + Angle ACB = 180^o
Produce BC to D. Through C draw CE || BA.

Proof of angle sum theorem of a Triangle:

Statement	Reason
1. Angle A = Angle ACE	Alternate angles angles BA is parallel to CE
2. Angle B = Angle ECD	Corresponding angles BA is parallel to CE
3. Angle A + angle B = Angle ACE + Angle ECD	statements (1) and (2)
4. Angle A + angle B  = Angle ACD	statement (3)
5. Angle A + Angle B + Angle ACB = Angle ACD + Angle   ACB	adding Angle ACB to both sides
6. But Angle ACD + Angle ACB = 180o	linear pair
7. Angle A + Angle B + Angle ACB = 180 °	statements (5) and (6)

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Corollary of Theorem on Angle theorem of a Triangle

If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles.
Given:
In Triangle ABC, BC is produced to D.
To Prove Corollary of Sum of Angles of Triangle:
Angle ACD = Angle A + Angle B



Proof:

Statement	Reason
1. Angle ACB + Angle ACD = 180o.	linear pair
2. Angle A + Angle B + Angle ACB = 180o	sum of the angles of a triangle = 180
3.  Angle ACB + Angle ACD = Angle A + Angle B + Angle ACB	statements (1) and (2)
4. Angle ACD = Angle A + Angle B Reason	statement (3); Angle ACB is common

Hard Fraction Problems

Hard fraction problems:

This articles discusses hard fraction problems solving. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)

Hard fraction problems solve for proper fraction, improper fraction and complex fraction problems.

Example for Hard fraction problems:

Example 1:
Subtract the fractions `4/5` – `3/4`

Solution:
The denominator (bottom number) is different so we have to take least common denominator (lcd).
LCD = 5 x 4 = 20
`(4 xx 4)/ (5 xx 4)` = `16/20` and `(3 xx 5)/ (4 xx 5)` = `15/20`
`16/20 ` – `15/20`
The denominators are equals
So subtracting the numerator directly = `(16-15)/20`
Simplify the above equation we get = `1/20`
Therefore the final answer is` 1/20`

Example 2:
Subtract the mixed fractions for given fractions,` 4 7/5` – `5 8/5`

Solution:
The given two mixed fractions are `4 7/5` – `5 7/5`
We need convert to mixed fraction to improper fraction `27/5` – `33/5`
The same denominators of the two fractions, so
                                         = `27/5` – `33/5`
Subtract the numerators the 27 and 33 = 27 - 33 = - 6.
The same denominator is 5.
                                         = `-6/5`
The subtract fraction solution is -`6/5` .

Example 3:
Multiply the mixed fractions for given two fraction,` 4 2/4` x `5 2/6`

Solution:
The given two mixed fractions are `4 2/4` x `5 2/6`
We need convert to mixed fraction to improper fraction `18/4` x `32/6`
Multiply the numerators the 18 and 32 = 18 x 32 = 576.
Multiply the denominators the 4 and 6 = 4 x 6 = 24
                                         =` 576/24`
The multiply fraction solution is 24

Example 4:
Convert `16/ (5/4)` to a simple fraction and reduce.

Solution:
The given complex fraction `16/ (5/4)`
Can be written as `16/1 -: 5/4`
First we have to take the reciprocal of the 2^nd number, and then multiply with the second one
Reciprocal of `5/4 ` is `4/5`
`16/1` x `4/5`
Multiply the numerator and denominator
`(16 xx 4) / (1 xx 5)` = `64/5`
Therefore complex fraction solution is `64/5`

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Practice for hard fraction problems:

Problem 1: Convert `5/ (5/4)` to a simple fraction and reduce.
Solution: 4.
Problem 2: Adding the mixed fractions for given two fraction,` 4 2/3` + `5 2/3`
Solution: `31/3`
Problem 3: Subtract the mixed fractions for given fractions,` 4 7/3` – `5 8/3`

Solution:`-4/3`

Written Translation

Introduction:
In word problem we can use the English words for translating words into mathematical expressions. In the translation we can use the math equation. This is used in the math problem to convert in the fairly simple problems. But we can’t do the figuring problem. This translating method is help to solve the word problems.

1. Working clearly will help you think clearly, and
2. Figuring out what you require will help you convert your final answer back into English.

Written translation - Translation methods:

• Addition
• Subtraction
• Multiplication
• Division
• Power

Written translation - Addition
Words that mean adding

words that mean adding	example
the sum of	the sum of a number and 6	y + 6
more than	6 more than a number	y + 6
plus	a number plus two	y + 2
added to	6 added to a number	y + 6
increased by	a number increased by 3	y + 3

Written translation - Subtraction

Words that mean subtracting

words that mean subtracting	example
less than	6 less than a number	y - 6
less	6 less a number	6 - y
minus	6 minus a number	6 - y
subtract from	subtract 6 from a number	y - 6
reduced by	a number reduced by 6	y - 6
decreased by	a number decreased by 6	y - 6
diminished by	6 diminished by a number	6 - y

Written translation - Multiplication

Words that mean multiplying

Words that mean multiplying	example
multiply	multiply a number by 3	3y
multiplied by	a number multiplied by 3	3y
times	4 times a number	4y
double	double a number	2y
twice	twice a number	2y
triple	triple a number	3y

Written translation - Division

Words that mean dividing

Words that mean dividing	example
divided by	a number divided by 6	y/6
divide into	divide 6 into a number	y/6
the quotient of	the quotient of a number and 3	y/3
half of	half of a number	y/2
one third of	one third of a number	y/3

Power

Word that mean rising to the n^th power

words that mean raising to the nth power	example
a number raised to the nth power	a number raised to the fifth power	y6
raise a number to the nth power	raise a number to the fifth power	y6
the nth power of a number	the fifth power of a number	y6
squared	a number squared	y2
cubed	a number cubed	y3

Binomial Theory

In algebra, a binomial theory is nothing but the study of binomial. The binomial is defined as the polynomial with sum of two monomial terms. Some of the binomial term is given as,The binomial also consist of distributions, coefficient, variables etc.some of the binomial terms are given as,

3x + 1 = 0

4x² + 5 = 0

9x³ + 15 = 0

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).



Properties of binomial:

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

A sum of binomial coefficients of even term is equal to a sum of binomial coefficients of odd terms, and it is equal to

2n-1

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

Examples

Example1: find the value of 5²- 3² and prove it?

Solution:

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).

52 − 32 = (5 + 3) (5 − 3).

= (8)(2)

= 16.      ---------- (1)

Proof:

a2 − b2 = 52 − 32

= 25 – 9

= 16       ---------- (2)

From 1 and 2 it proved.

Example 2: find the value of (3x + 1) and (2x + 3)?

Solution:

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

(3x + 1)(2x + 3) = 3*2x2 + 3*3x + 1*2x + 1*3.

= 6x2 + 9x + 2x + 3.

Therefore, (3x + 1) (2x + 3) = 6x2 + 9x + 2x + 3

Example 3: Prove that (a + b) ³ = 8?

Solution:

From the property of binomial,

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

(a + b) ³ = (1 + 1) 3 = 2 3

= 8

Hence it is proved.

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Practice problem:

Problem1: find the value of 9²- 4² and prove it?

Answer is 65

Problem2: find the value of (2x + 1) and (x + 3)?

Answer is 2x2 + 6x + 1x + 3.

Solving Sums Integrals

Integration is an important concept in mathematics and, together with differentiation, is one of the two main operations in calculus. The term integral may also refer to the notion of antiderivative, a function F whose derivative is the given function f. In this case it is called an indefinite integral. Integral can be classified as definite and indefinite integral. (Source: Wikipedia)

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Examples problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function ∫ (234x2 + 132x4 - 5x) dx.

Solution:

Given ∫ (234x2 + 132x4 - 5x) dx

Integrate the given function with respect to x, we get

∫ (234x2 + 132x4 - 5x) dx = ∫ 234x2 dx + ∫ 132x4 dx - ∫ 5x dx.

= 234 (x3 / 3) + 132 (x5 / 5) - 5 (x2 / 2) + c.

= 78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Answer:

The final solution is  78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Solving sums integral problem 2`:`

Find the value of the integration

`int_2^5(x^6)dx`

Solution:

Integrate the given function with respect to x, we get

`int_2^5(x^6)dx`  = `(x^7 / 7)`52

Substitute the lower and upper limits, we get

= `((5^7 / 7) - (2^7/ 7))`

= `((78125 / 7) - (128 / 7))`

= `(77997 / 7)`

Answer:

The final answer is `(77997 / 7)`

Solving sums integral problem 3:

Integration using algebraic rational function ∫ 7dx / (17x + 37)

Solution:

Using integrable function method,

Given function is ∫ `(7dx) / (17x + 37)`

Formula:

∫ [L / (ax + c)] dx = (L / a) log (ax + c)

From given, L = 7, a = 17, and c = 37

Integrate the given equation with respect to x, we get

= `(7 / 17)` log (17x + 37)

Answer:

The final answer is `(7 / 17)` log (17x + 37).

Practice problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function using integrable function ∫ `(14 / (11x + 12))` dx

Answer:

The final answer is `(14 / 11)` log (11x +12)

Solving sums integral problem 2:

Integrate the given function using integrable function ∫ `(15 / (21x + 92))` dx

Answer:

The final answer is `(15 / 21)` log (21x + 92)

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Solving sums integral problem 3:

Integrate the given function ∫ (7.9x2 - 12.9x) dx

Answer:

The final answer is `(7.9 / 3)` x3 - `(12.9 / 2)` x2

Difference of Two Means

Mean is defined as the average for the total number of values given in the data set. It is the sum made between the given data set and it is divided it by the total number of values given in the data set. Tis is called as the mean for the given data set. This can also be called as arithametic mean or sample mean. The difference between two mean is nothing but the taking difference for the two sets of data (mu_1 , mu_2 ) and calculating the difference for that mean.

Difference of two mean  = mu_1 - mu_2

Where as

mu_1 is the mean value for the first set of data.
mu_2 is the mean value for the second set of data.

Mean is calculated by the way,
mu = (Sum of all the values given) / (Total Number of values)


Steps for calculating the difference of two means:

Get the two sets of data for calculating the mean.
Measure the sum for the first set of data and divide it by the total number of data's given. This is the mean for the first data set
Measure the sum for the second set of data and divide it by the total number of data's given. This is the mean for the second data set.
Now measure the difference of the two means from the mean calculated.



Difference of two means - Example Problems:

Difference of two means - Problem 1:

Find the difference of two means from the given two data set. 2, 3, 4, 5, 6, 7 and 1, 2, 3, 4, 5, 6

Solution:

Mean for the first set of the data given

mu_1 = (2+3+4+5+6+7) / 6

= 27 / 6

=  4.5

Mean for the second set of the data given

mu_2 = (1+2+3+4+5+6) / 6

=  21 / 6

mu_2    = 3.5

Difference of the two means is given by

Difference of two mean = mu_1 - mu_2

= 4.5 - 3.5

Difference of two mean = 1

Difference of two means - Problem 2:

Find the difference of two means from the given two data set. 24, 23, 24, 25, 26, 27 and 11, 13, 13, 14, 15, 16

Solution:

Mean for the first set of the data given

mu_1 = (24+23+24+25+26+27) / 6

= 149 / 6

=  24.8333333

Mean for the second set of the data given

mu_2 = (11+13+13+14+15+16) / 6

=  82 / 6

mu_2    = 13.6666667

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Difference of the two means is given by

Difference of two mean = mu_1 - mu_2

= 24.8333333 - 13.6666667

Difference of two mean  = 11.1666666